Commission Directive 2005/78/EC of 14 November 2005 implementing Directive 2005/55/EC of the European Parliament and of the Council on the approximation of the laws of the Member States relating to the measures to be taken against the emission of gaseous and particulate pollutants from compression-ignition engines for use in vehicles, and the emission of gaseous pollutants from positive ignition engines fuelled with natural gas or liquefied petroleum gas for use in vehicles and amending Annexes I, II, III, IV and VI thereto (Text with EEA relevance) (repealed)
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The determination of system equivalency according to section 6.2 of this Annex shall be based on a 7 sample pair (or larger) correlation study between the candidate system and one of the accepted reference systems of this Directive using the appropriate test cycle(s). The equivalency criteria to be applied shall be the F-test and the two-sided Student t-test.
This statistical method examines the hypothesis that the population standard deviation and mean value for an emission measured with the candidate system do not differ from the standard deviation and population mean value for that emission measured with the reference system. The hypothesis shall be tested on the basis of a 5 % significance level of the F and t values. The critical F and t values for 7 to 10 sample pairs are given in the table below. If the F and t values calculated according to the formulae below are greater than the critical F and t values, the candidate system is not equivalent.
The following procedure shall be followed. The subscripts R and C refer to the reference and candidate system, respectively:
Conduct at least 7 tests with the candidate and reference systems preferably operated in parallel. The number of tests is referred to as nR and nC.
Calculate the mean values xR and xC and the standard deviations sR and sC.
Calculate the F value, as follows:
(the greater of the two standard deviations SR or SC must be in the numerator)
Compare the calculated F and t values with the critical F and t values corresponding to the respective number of tests indicated in table below. If larger sample sizes are selected, consult statistical tables for 5 % significance (95 % confidence) level.
Determine the degrees of freedom (df), as follows:
:
df = nR – 1 / nC – 1
:
df = nC + nR – 2
| F and t values for selected sample sizes | ||||
| Sample Size | F-test | t-test | ||
|---|---|---|---|---|
| df | Fcrit | df | tcrit | |
| 7 | 6/6 | 4,284 | 12 | 2,179 |
| 8 | 7/7 | 3,787 | 14 | 2,145 |
| 9 | 8/8 | 3,438 | 16 | 2,120 |
| 10 | 9/9 | 3,179 | 18 | 2,101 |
Determine the equivalency, as follows:
if F < Fcrit and t < tcrit, then the candidate system is equivalent to the reference system of this Directive,
if F ≥ Fcrit and t ≥ tcrit, then the candidate system is different from the reference system of this Directive.
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